Hydrated copper sulfate is heated. Using the speculative chemical equation CuSO4*5H2O-->CuSo4+5H2O, show how you could have predicted, by calculations, what the final weights in the crucibles might be. Show calculations for two starting weights between 1 and 4 grams.
1 g CuSO4.5H2O x (molar mass CuSO4/CuSO4.5H2O) = ? wt CuSO4.
Starting weight = wt xble + 1g
Ending weight = wt xble + ?wt CuSO4
To predict the final weights of the products in the crucibles, we need to first understand the stoichiometry of the chemical reaction and then perform the necessary calculations. The given chemical equation shows that hydrated copper sulfate (CuSO4·5H2O) is heated, producing anhydrous copper sulfate (CuSO4) and water (H2O) as products.
Let's use the molar masses of the compounds involved to perform the calculations:
- Molar mass of CuSO4·5H2O:
- Cu: 63.55 g/mol
- S: 32.07 g/mol
- O: 16.00 g/mol (x4)
- H: 1.01 g/mol (x10)
- Total molar mass = (63.55 + 32.07 + 16.00*4 + 1.01*10) g/mol = 249.68 g/mol
- Molar mass of CuSO4:
- Cu: 63.55 g/mol
- S: 32.07 g/mol
- O: 16.00 g/mol (x4)
- Total molar mass = (63.55 + 32.07 + 16.00*4) g/mol = 159.61 g/mol
- Molar mass of H2O:
- H: 1.01 g/mol (x2)
- O: 16.00 g/mol
- Total molar mass = (1.01*2 + 16.00) g/mol = 18.02 g/mol
Now, let's perform the calculations for two different starting weights between 1 and 4 grams. We'll assume that the crucible is completely empty before adding the hydrated copper sulfate.
Starting weight 1:
- Assume we have 1.5 grams of CuSO4·5H2O in the crucible.
- Calculate the number of moles of CuSO4·5H2O:
- Moles = Mass / Molar mass = 1.5 g / 249.68 g/mol ≈ 0.006 mol
- According to the balanced equation, the molar ratio between CuSO4·5H2O and CuSO4 is 1:1. Therefore, the number of moles of CuSO4 produced is also 0.006 mol.
- Calculate the mass of CuSO4 produced:
- Mass = Moles × Molar mass = 0.006 mol × 159.61 g/mol ≈ 0.96 g
- According to the balanced equation, the molar ratio between CuSO4·5H2O and H2O is 1:5. Therefore, the number of moles of H2O produced is 5 times the moles of CuSO4·5H2O.
- Calculate the mass of H2O produced:
- Mass = Moles × Molar mass = 0.006 mol × 18.02 g/mol ≈ 0.11 g
- The final weight of CuSO4 in the crucible will be 0.96 g, and the final weight of H2O will be 0.11 g.
Starting weight 2:
- Assume we have 3.2 grams of CuSO4·5H2O in the crucible.
- Calculate the number of moles of CuSO4·5H2O:
- Moles = Mass / Molar mass = 3.2 g / 249.68 g/mol ≈ 0.013 mol
- According to the balanced equation, the molar ratio between CuSO4·5H2O and CuSO4 is 1:1. Therefore, the number of moles of CuSO4 produced is also 0.013 mol.
- Calculate the mass of CuSO4 produced:
- Mass = Moles × Molar mass = 0.013 mol × 159.61 g/mol ≈ 2.08 g
- According to the balanced equation, the molar ratio between CuSO4·5H2O and H2O is 1:5. Therefore, the number of moles of H2O produced is 5 times the moles of CuSO4·5H2O.
- Calculate the mass of H2O produced:
- Mass = Moles × Molar mass = 0.013 mol × 18.02 g/mol ≈ 0.23 g
- The final weight of CuSO4 in the crucible will be 2.08 g, and the final weight of H2O will be 0.23 g.
So, using the given chemical equation and performing the calculations as described above, we can predict the final weights of CuSO4 and H2O in the crucibles for two different starting weights between 1 and 4 grams.